CH6901 骑士放置

原题链接

和棋盘覆盖(题解)差不多.。
同样对格子染色，显然日字的对角格子是不同色，直接在对应节点连边，然后就是二分图最大独立集问题。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1e4 + 10;
const int M = 2.6e7 + 10;
int fi[N], di[M], ne[M], mtc[N], mo_x[8] = { -1, -2, -2, -1, 1, 2, 2, 1 }, mo_y[8] = { -2, -1, 1, 2, -2, -1, 1, 2 }, l, n, m;
bool v[N], a[102][102];
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c < '0' || c > '9'; c = getchar())
		p |= c == '-';
	for (; c >= '0' && c <= '9'; c = getchar())
		x = x * 10 + c - '0';
	return p ? -x : x;
}
inline void add(int x, int y)
{
	di[++l] = y;
	ne[l] = fi[x];
	fi[x] = l;
}
inline int ch(int x, int y)
{
	return (x - 1) * m + y;
}
bool dfs(int x)
{
	int i, y;
	for (i = fi[x]; i; i = ne[i])
		if (!v[y = di[i]])
		{
			v[y] = 1;
			if (!mtc[y] || dfs(mtc[y]))
			{
				mtc[y] = x;
				return true;
			}
		}
	return false;
}
int main()
{
	int i, x, y, j, s = 0, k, o;
	n = re();
	m = re();
	k = re();
	for (i = 1; i <= k; i++)
	{
		x = re();
		y = re();
		a[x][y] = 1;
	}
	for (i = 1; i <= n; i++)
		for (j = 1; j <= m; j++)
			if (!a[i][j] && !((i + j) & 1))
				for (o = 0; o < 8; o++)
				{
					x = i + mo_x[o];
					y = j + mo_y[o];
					if (x > 0 && x <= n && y > 0 && y <= m && !a[x][y])
						add(ch(i, j), ch(x, y));
				}
	for (i = 1; i <= n; i++)
		for (j = 1; j <= m; j++)
			if (!a[i][j] && !((i + j) & 1))
			{
				memset(v, 0, sizeof(v));
				if (dfs(ch(i, j)))
					s++;
			}
	printf("%d", n * m - s - k);
	return 0;
}